Question

1g of carbonate of metal was dissolved in 25 ml of 1N HCl. The resulting liquid required 5 ml of 1N NaOH for neutralization. The equivalent weight of the metal carbonate is:-

50

30

20

None of these

Answer 1

For 5ml of 1 N NaOH, HCl required is:
M1V1 = M2V2
1*5 = 1*V2
V2 = 5 ml
So, HCl required will be 5 ml.
So, volume of HCl left = 25-5 = 20 ml = 0.020 L
Concentration of HCl = 1 N
For HCl, 1N = 1 M, because n-factor = 1
molarity = number of moles/volume
1 M = ​number of moles/0.020
number of moles = 0.020 mol
0.020 mol of HCl will be used in neutralization of carbonate.
The reaction is:
M2CO3 + 2HCl ----> 2MCl + H2CO3
1 mol of M2CO3 is neutralized by 2 moles of HCl. If 0.020 mol of HCl is used, then M2CO3 used = 0.010 mol
mass of M2CO3 = 1 g
moles = 0.010 mol
molar mass = mass/moles = 1/0.010 = 100 g
Molar mass = 100 g/mol
n-factor = 2 (for carbonate)
equivalent weight = 100/2 = 50
Equivalent weight of metal carbonate = 50
correct option is 1.

Answer 2

Answer:

The equivalent weight of the metal carbonate is 50.

Explanation:

Given:

Weight of metal carbonate=1g

The volume of 1N HCl=25 ml

The volume of NaOH=5 ml

To find:

The equivalent weight of the metal carbonate=?
Solution:
The volume of HCl used=25-5=20ml

∴ mEq of HCl = mEq of carbonate

$\frac{20}{1000}Eq=\frac{Weight}{Eq} carbonate=\frac{1}{Ew}$

∴Ew=1000/20=50

Therefore, the equivalent weight of the metal carbonate is 50.